package algorithm.problems.depth_first_search;

import java.util.*;

/**
 * Created by gouthamvidyapradhan on 27/12/2017.
 * Given a list accounts, each element accounts[i] is a list of strings, where the first element accounts[i][0] is a
 * name, and the rest of the elements are emails representing emails of the account.

 Now, we would like to merge these accounts. Two accounts definitely belong to the same person if there is some email
 that is common to both accounts. Note that even if two accounts have the same name, they may belong to different
 people  as people could have the same name. A person can have any number of accounts initially, but all of their
 accounts definitely have the same name.

 After merging the accounts, return the accounts in the following format: the first element of each account is the
 name,  and the rest of the elements are emails in sorted order. The accounts themselves can be returned in any order.

 Example 1:
 Input:
 accounts = [["John", "johnsmith@mail.com", "john00@mail.com"], ["John", "johnnybravo@mail.com"], ["John",
 "johnsmith@mail.com", "john_newyork@mail.com"], ["Mary", "mary@mail.com"]]
 Output: [["John", 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com'],  ["John",  "johnnybravo@mail
 .com"], ["Mary", "mary@mail.com"]]
 Explanation:
 The first and third John's are the same person as they have the common email "johnsmith@mail.com".
 The second John and Mary are different people as none of their email addresses are used by other accounts.
 We could return these lists in any order, for example the answer [['Mary', 'mary@mail.com'], ['John',
 'johnnybravo@mail.com'],
 ['John', 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com']] would still be accepted.
 Note:

 The length of accounts will be in the range [1, 1000].
 The length of accounts[i] will be in the range [1, 10].
 The length of accounts[i][j] will be in the range [1, 30].

 Solution: Consider each email_address as a vertex and link the emails of each account as bidirectional edges and
 perform a dfs on connected components and return the email addresses of connected components.
 */
public class AccountsMerge {

    private class Account{
        String email, name;
        Account(String email, String name){
            this.email = email;
            this.name = name;
        }
    }

    private Map<Integer, Account> numMap; //for simplicity mapping each email to a unique integer
    private Map<String, Integer> emailMap;
    private Map<Integer, List<Integer>> graph;
    private BitSet done;
    /**
     * Main method
     * @param args
     * @throws Exception
     */
    public static void main(String[] args) throws Exception{
        List<String> account1 = new ArrayList<>();
        List<List<String>> accounts = new ArrayList<>();
        account1.add("John");
        account1.add("johnsmith@mail.com");
        account1.add("john00@mail.com");
        accounts.add(account1);

        List<String> account2 = new ArrayList<>();
        account2.add("John");
        account2.add("johnnybravo@mail.com");
        accounts.add(account2);

        List<String> account3 = new ArrayList<>();
        account3.add("John");
        account3.add("johnsmith@mail.com");
        account3.add("john_newyork@mail.com");
        accounts.add(account3);

        List<String> account4 = new ArrayList<>();
        account4.add("Mary");
        account4.add("mary@mail.com");
        accounts.add(account4);

        List<List<String>> result = new AccountsMerge().accountsMerge(accounts);
        System.out.println(result);
    }

    public List<List<String>> accountsMerge(List<List<String>> accounts) {
        done = new BitSet();
        numMap = new HashMap<>();
        emailMap = new HashMap<>();
        graph = new HashMap<>();
        int index = -1;
        for(List<String> account : accounts){
            String name = account.get(0);
            int prev = -1;
            for(int i = 1, l = account.size(); i < l; i ++){
                String email = account.get(i);
                int vertex;
                if(!emailMap.containsKey(email)){
                    vertex = ++index;
                    emailMap.put(email, vertex);
                    numMap.put(vertex, new Account(email, name));
                } else {
                    vertex = emailMap.get(email);
                }
                graph.putIfAbsent(vertex, new ArrayList<>());
                if(i != 1){
                    //make bi-directional link
                    graph.get(prev).add(vertex);
                    graph.get(vertex).add(prev);
                }
                prev = vertex;
            }
        }
        List<List<String>> result = new ArrayList<>();
        for(int i : graph.keySet()){
            if(!done.get(i)){
                List<Integer> list = new ArrayList<>();
                List<String> account = new ArrayList<>();
                dfs(i, list);
                list.stream().forEach(x -> account.add(numMap.get(x).email));
                account.sort(String::compareTo);

                //Add account user name
                Account acc = numMap.get(list.get(0));
                account.add(0, acc.name);
                result.add(account);
            }
        }
        return result;
    }

    private void dfs(int i, List<Integer> list){
        done.set(i);
        list.add(i);
        List<Integer> children = graph.get(i);
        if(children != null){
            for(int c : children){
                if(!done.get(c)){
                    dfs(c, list);
                }
            }
        }
    }

}
